Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 4} = \dfrac{-2x + 24}{x - 4}$
Multiply both sides by $x - 4$ $ \dfrac{x^2}{x - 4} (x - 4) = \dfrac{-2x + 24}{x - 4} (x - 4)$ $ x^2 = -2x + 24$ Subtract $-2x + 24$ from both sides: $ x^2 - (-2x + 24) = -2x + 24 - (-2x + 24)$ $ x^2 + 2x - 24 = 0$ Factor the expression: $ (x - 4)(x + 6) = 0$ Therefore $x = 4$ or $x = -6$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.